# 6.5 De broglie’s matter waves  (Page 2/10)

 Page 2 / 10
$\lambda =\frac{h}{p}=\frac{hc}{\sqrt{K\left(K+2{E}_{0}\right)}}.$

Depending on the problem at hand, in this equation we can use the following values for hc : $hc=\left(6.626\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}\right)\left(2.998\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{m/s}\right)=1.986\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-25}\text{J}·\text{m}=1.241\phantom{\rule{0.2em}{0ex}}\text{eV}·\mu \text{m}$

## Solution

1. For the basketball, the kinetic energy is
$K={m}_{0}{u}^{2}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}2=\left(0.65\text{kg}\right){\left(10\text{m/s}\right)}^{2}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}2=32.5\text{J}$

and the rest mass energy is
${E}_{0}={m}_{0}{c}^{2}=\left(0.65\text{kg}\right){\left(2.998\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{m/s}\right)}^{2}=5.84\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{16}\text{J.}$

We see that $K\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\left(K+{E}_{0}\right)\ll 1\phantom{\rule{0.2em}{0ex}}\text{and use}\phantom{\rule{0.2em}{0ex}}p={m}_{0}u=\left(0.65\text{kg}\right)\left(10\text{m/s}\right)=6.5\phantom{\rule{0.2em}{0ex}}\text{J}·\text{s/m}:$
$\lambda =\frac{h}{p}=\frac{6.626\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}}{6.5\text{J}·\text{s/m}}=1.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{m}.$
2. For the nonrelativistic electron,
${E}_{0}={m}_{0}{c}^{2}=\left(9.109\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\text{kg}\right){\left(2.998\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\text{m/s}\right)}^{2}=511\phantom{\rule{0.2em}{0ex}}\text{keV}$

and when $K=1.0\phantom{\rule{0.2em}{0ex}}\text{eV},$ we have $K\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\left(K+{E}_{0}\right)=\left(1\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}512\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\ll 1,$ so we can use the nonrelativistic formula. However, it is simpler here to use [link] :
$\lambda =\frac{h}{p}=\frac{hc}{\sqrt{K\left(K+2{E}_{0}\right)}}=\frac{1.241\phantom{\rule{0.2em}{0ex}}\text{eV}·\mu \text{m}}{\sqrt{\left(1.0\phantom{\rule{0.2em}{0ex}}\text{eV}\right)\left[1.0\phantom{\rule{0.2em}{0ex}}\text{eV+}2\left(511\phantom{\rule{0.2em}{0ex}}\text{keV}\right)\right]}}=1.23\phantom{\rule{0.2em}{0ex}}\text{nm}.$

If we use nonrelativistic momentum, we obtain the same result because 1 eV is much smaller than the rest mass of the electron.
3. For a fast electron with $K=108\phantom{\rule{0.2em}{0ex}}\text{keV,}$ relativistic effects cannot be neglected because its total energy is $E=K+{E}_{0}=108\phantom{\rule{0.2em}{0ex}}\text{keV}+511\phantom{\rule{0.2em}{0ex}}\text{keV}=619\phantom{\rule{0.2em}{0ex}}\text{keV}$ and $K\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}E=108\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}619$ is not negligible:
$\lambda =\frac{h}{p}=\frac{hc}{\sqrt{K\left(K+2{E}_{0}\right)}}=\frac{1.241\phantom{\rule{0.2em}{0ex}}\text{eV}·\text{μm}}{\sqrt{108\phantom{\rule{0.2em}{0ex}}\text{keV}\left[108\phantom{\rule{0.2em}{0ex}}\text{keV}+2\left(511\phantom{\rule{0.2em}{0ex}}\text{keV}\right)\right]}}=3.55\phantom{\rule{0.2em}{0ex}}\text{pm}.$

## Significance

We see from these estimates that De Broglie’s wavelengths of macroscopic objects such as a ball are immeasurably small. Therefore, even if they exist, they are not detectable and do not affect the motion of macroscopic objects.

Check Your Understanding What is de Broglie’s wavelength of a nonrelativistic proton with a kinetic energy of 1.0 eV?

1.7 pm

Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron’s angular momentum in the hydrogen atom, which was postulated in Bohr’s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave. To see it clearly, imagine a stretched guitar string that is clamped at both ends and vibrates in one of its normal modes. If the length of the string is l ( [link] ), the wavelengths of these vibrations cannot be arbitrary but must be such that an integer k number of half-wavelengths $\lambda \phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}2$ fit exactly on the distance l between the ends. This is the condition $l=k\lambda \phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}2$ for a standing wave on a string. Now suppose that instead of having the string clamped at the walls, we bend its length into a circle and fasten its ends to each other. This produces a circular string that vibrates in normal modes, satisfying the same standing-wave condition, but the number of half-wavelengths must now be an even number $k,\phantom{\rule{0.2em}{0ex}}k=2n,$ and the length l is now connected to the radius ${r}_{n}$ of the circle. This means that the radii are not arbitrary but must satisfy the following standing-wave condition:

$2\pi {r}_{n}=2n\frac{\lambda }{2}.$

If an electron in the n th Bohr orbit moves as a wave, by [link] its wavelength must be equal to $\lambda =2\pi {r}_{n}\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}n.$ Assuming that [link] is valid, the electron wave of this wavelength corresponds to the electron’s linear momentum, $p=h\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\lambda =nh\phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}\left(2\pi {r}_{n}\right)=n\hslash \phantom{\rule{0.1em}{0ex}}\text{/}\phantom{\rule{0.1em}{0ex}}{r}_{n}.$ In a circular orbit, therefore, the electron’s angular momentum must be

${L}_{n}={r}_{n}p={r}_{n}\frac{n\hslash }{{r}_{n}}=n\hslash .$

This equation is the first of Bohr’s quantization conditions, given by [link] . Providing a physical explanation for Bohr’s quantization condition is a convincing theoretical argument for the existence of matter waves.

fundamental note of a vibrating string
what are matter waves? Give some examples
according to de Broglie any matter particles by attaining the higher velocity as compared to light'ill show the wave nature and equation of wave will applicable on it but in practical life people see it is impossible however it is practicaly true and possible while looking at the earth matter at far
Manikant
Mathematical expression of principle of relativity
given that the velocity v of wave depends on the tension f in the spring, it's length 'I' and it's mass 'm'. derive using dimension the equation of the wave
What is the importance of de-broglie's wavelength?
he related wave to matter
Zahid
at subatomic level wave and matter are associated. this refering to mass energy equivalence
Zahid
it is key of quantum
Manikant
how those weight effect a stable motion at equilibrium
how do I differentiate this equation- A sinwt with respect to t
just use the chain rule : let u =wt , the dy/dt = dy/du × du/dt : wA × cos(wt)
Jerry
I see my message got garbled , anyway use the chain rule with u= wt , etc...
Jerry
de broglie wave equation
vy beautiful equation
chandrasekhar
what is electro statics
when you consider systems consisting of fixed charges
Sherly
Diagram of the derive rotational analog equation of v= u+at
what is carat
a unit of weight for precious stones and pearls, now equivalent to 200 milligrams.
LoNE
a science that deals with the composition, structure, and properties of substances and with the transformations that they undergo.
LoNE
what is chemistry
what chemistry ?
Abakar
where are the mcq
ok
Giorgi
acids and bases
Navya
How does unpolarized light have electric vector randomly oriented in all directions.
unpolarized light refers to a wave collection which has an equal distribution of electric field orientations for all directions
pro
In a grating, the angle of diffraction for second order maximum is 30°.When light of wavelength 5*10^-10cm is used. Calculate the number of lines per cm of the grating.
OK I can solve that for you using Bragg's equation 2dsin0over lander
ossy